The polynomial p{x) = x4 -2x3 + 3x2 -ax+3a-7 when divided by x+1 leaves the remainder 19. Find the values of a. Also, find the remainder when p(x) is divided by x+ 2.
Given, $\quad p(x)=x^{4}-2 x^{3}+3 x^{2}-a x+3 a-7$
When we divide $p(x)$ by $x+1$, then we get the remainder $p(-1)$.
Now, $\quad p(-1)=(-1)^{4}-2(-1)^{3}+3(-1)^{2}-a(-1)+3 a-7$
$=1+2+3+a+3 a-7=4 a-1$
According to the question, $p(-1)=19$
$\Rightarrow \quad 4 a-1=19$
$\Rightarrow \quad 4 a=20$
$\therefore \quad a=5$
$\therefore$ Required polynomial $=x^{4}-2 x^{3}+3 x^{2}-5 x+3(5)-7$ [put $a=5$ in $p(x)$ ]
$=x^{4}-2 x^{3}+3 x^{2}-5 x+15-7$
$=x^{4}-2 x^{3}+3 x^{2}-5 x+8$
When we divide $p(x)$ by $x+2$, then we get the remainder $p(-2)$.
Now, $\quad p(-2)=(-2)^{4}-2(-2)^{3}+3(-2)^{2}-5(-2)+8$
$=16+16+12+10+8=62$
Hence, the value of $a$ is 5 and remainder is 62 .